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lord [1]
3 years ago
13

The enthalpy change for the explosion of ammonium nitrate with fuel oil is –7198 kJ for every 3 moles of NH4NO3. What is the ent

halpy change for 1.0 mole of NH4NO3 in this reaction?
Chemistry
1 answer:
anastassius [24]3 years ago
7 0

Answer:

−2399.33 kJ

Explanation:

If NH₄NO₃ reacts with fuel oil to give a ΔH of -7198 for every 3 moles of NH₄NO₃

What is the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction

∴ For every 1 mole, we will have \frac{1}{3} of the total enthaply of the 3 moles

so, to determine the 1 mole; we have:

\frac{1}{3}*(-7198kJ)

= −2399.33 kJ

∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ

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How many km are in 5.6mm? 5.6x103 5.6x10-6 5.6x10-3 5.6x106​
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Hey there :)

<em>Q</em><em>u</em><em>e</em><em>s</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em> </em><em>How many km are in 5.6mm? </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>3 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-6 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-3 </em>

<em>=</em><em>></em><em> </em><em>5.6x10</em><em>^</em><em>6</em>

<em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em>-</em>

5.6 \times 10^{ - 6}

<em>E</em><em>x</em><em>p</em><em>l</em><em>a</em><em>n</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em>-</em>

By using the formula-

1millimeter =  \frac{1}{1000000}

As 1 with 6 zeros, we convert it into exponential form.

=  >  \frac{1}{10^{6} }

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

=  > 10^{ - 6}

Now combine with given question.

=  > 5.6 \times 10^{ - 6}

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